Current Electricity — Class 12 Physics practice questions

Question 1

A technician is testing two resistors from a hardware store. Resistor P has a length of 2 m, cross-sectional area 1×10⁻⁶ m², and resistivity 1×10⁻⁶ Ω m. Resistor Q is made of the same material but has double the length and half the cross-sectional area of P.

(i) Calculate the resistance of resistor P.
(ii) Find the ratio of resistance of Q to resistance of P.
(iii) The two resistors P and Q are connected in parallel across a 6 V battery of negligible internal resistance. Find the total current drawn from the battery.
(iv) If the temperature of resistor P is increased, state and explain how its resistance changes, given it is a metallic conductor.

Accepted Answer

(i) Resistance of resistor P:

Using the formula R = ρL/A,

R_P = (1×10⁻⁶ × 2) / (1×10⁻⁶)

∴ R_P = 2 Ω

(ii) Ratio of resistance of Q to resistance of P:

For resistor Q: length L_Q = 2L = 4 m, area A_Q = A/2 = 0.5×10⁻⁶ m², same ρ.

R_Q = ρ × (2L) / (A/2) = ρ × 4L/A = 4 × (ρL/A) = 4 R_P

∴ R_Q / R_P = 4 (i.e., R_Q = 8 Ω)

(iii) Total current from the battery:

P and Q are in parallel across V = 6 V.

Current through P: I_P = V/R_P = 6/2 = 3 A

Current through Q: I_Q = V/R_Q = 6/8 = 0.75 A

By Kirchhoff's Current Law (KCL), total current I = I_P + I_Q = 3 + 0.75

∴ I = 3.75 A

(iv) Effect of temperature on resistance of metallic resistor P:

For metallic conductors, resistivity increases with temperature: ρ = ρ₀(1 + αΔT), where α > 0 for metals.

Since R = ρL/A and L, A remain practically unchanged, resistance R_P increases as temperature rises.

This is because increasing temperature causes more frequent collisions between free electrons and the vibrating lattice ions, reducing the relaxation time (τ), hence increasing resistivity and resistance.

Question 2

A materials testing laboratory uses a long cylindrical resistor of length 80 cm and cross-sectional area 2.0 × 10⁻⁶ m² to calibrate precision instruments. The resistor is made of nichrome (resistivity ρ = 1.0 × 10⁻⁶ Ω m). During a test, it is connected to a battery of EMF 6 V and internal resistance 1 Ω. Midway along its length, a thin wire of negligible resistance is tapped off, dividing the resistor into two equal halves. These two halves are reconnected in parallel between the battery terminals (internal resistance still 1 Ω, EMF still 6 V).

(i) Calculate the resistance of the full cylindrical resistor.
(ii) Find the equivalent external resistance when the two halves are connected in parallel.
(iii) Calculate the current drawn from the battery in the parallel arrangement.
(iv) A technician claims that connecting the two halves in parallel (instead of using the full resistor in series) will double the power dissipated in the external circuit. Verify whether this claim is correct, showing your calculation.

Accepted Answer

▌ Sub-part (i) — Resistance of full resistor [1 mark]

By the relation R = ρL/A,

R = (1.0 × 10⁻⁶ × 0.80) / (2.0 × 10⁻⁶)

∴ R = 0.40 Ω

▌ Sub-part (ii) — Equivalent resistance of two halves in parallel [1 mark]

Each half has resistance R/2 = 0.40/2 = 0.20 Ω.

For two equal resistors R₁ = R₂ = 0.20 Ω in parallel:

1/R_eq = 1/0.20 + 1/0.20 = 10

∴ R_eq = 0.10 Ω

▌ Sub-part (iii) — Current drawn from battery in parallel arrangement [1 mark]

By Ohm's law, total EMF drives current through (R_eq + r):

I = E / (R_eq + r) = 6 / (0.10 + 1) = 6 / 1.10

∴ I ≈ 5.45 A

▌ Sub-part (iv) — Verification of technician's claim [1 mark]

Power dissipated in external resistance P = I²R_ext.

Case 1 (full resistor R = 0.40 Ω in series with r = 1 Ω):
I₁ = 6 / (0.40 + 1) = 6 / 1.40 ≈ 4.286 A
P₁ = I₁² × R = (4.286)² × 0.40 ≈ 18.37 × 0.40 ≈ 7.35 W

Case 2 (parallel halves R_eq = 0.10 Ω):
I₂ = 5.45 A (from part iii)
P₂ = I₂² × R_eq = (5.45)² × 0.10 ≈ 29.70 × 0.10 ≈ 2.97 W

P₂ / P₁ ≈ 2.97 / 7.35 ≈ 0.40

∴ The power dissipated in the external circuit actually DECREASES (by a factor ≈ 0.40), not doubles. The technician's claim is incorrect. Although R_eq is reduced (drawing more current from the battery), the dominant effect is that the very low R_eq means most of the terminal voltage is dropped across the internal resistance, so external power is less.

Question 3

A technician tests a battery pack used in a solar lantern. In an open circuit, the terminal voltage of the battery reads 6.0 V. When the lantern (resistance 2 Ω) is switched ON, the terminal voltage drops to 5.0 V.

(i) Find the internal resistance of the battery.
(ii) Find the current drawn by the lantern when it is ON.
(iii) Calculate the power dissipated inside the battery due to its internal resistance.
(iv) If the same battery is accidentally short-circuited (R = 0), what will be the short-circuit current? Comment on why this is dangerous.

Accepted Answer

(i) Finding the internal resistance:

By the terminal voltage relation: ε = V + Ir

In open circuit, terminal voltage = EMF:
∴ ε = 6.0 V

When lantern is ON, terminal voltage V = 5.0 V

First, find current using Ohm's law for external resistance:
I = V/R = 5.0/2 = 2.5 A

Substituting in ε = V + Ir:
6.0 = 5.0 + 2.5 × r
→ r = 1.0/2.5

∴ r = 0.4 Ω

(ii) Finding the current drawn by the lantern:

Using I = V/R (already calculated above):
I = 5.0/2

∴ I = 2.5 A

(iii) Power dissipated inside the battery:

Using P = I²r:
P = (2.5)² × 0.4
P = 6.25 × 0.4

∴ P = 2.5 W

This energy is lost as heat inside the battery, reducing the efficiency of the lantern.

(iv) Short-circuit current (R = 0):

Using I_sc = ε/r (when R = 0, entire EMF drives current through internal resistance):
I_sc = 6.0/0.4

∴ I_sc = 15 A

This is dangerous because: the short-circuit current (15 A) is 6 times the normal operating current (2.5 A). Such a large current causes excessive heating (P = I²r = 225 × 0.4 = 90 W) inside the battery, which can cause the battery to overheat, swell, leak, or even catch fire, permanently damaging the battery pack.

Question 4

Assertion (A) : When the temperature of a metallic conductor increases, its electrical resistance increases.
Reason (R) : With rise in temperature, the number density of free electrons in a metal decreases.

Accepted Answer

Option (c) is correct.

Explanation: Assertion (A) is TRUE. For metals, resistivity varies as ρ = ρ₀(1 + αΔT), where α > 0. Hence resistance R = ρL/A increases with rising temperature because increased thermal agitation causes more frequent collisions of free electrons with lattice ions, reducing drift velocity.

Reason (R) is FALSE. The number density (n) of free electrons in a metal remains essentially constant with temperature; it is the relaxation time (τ) that decreases with increasing temperature, thereby increasing resistivity. Since R = ρL/A and ρ = m/ne²τ, a decrease in τ directly increases ρ and hence R.

∴ A is true, R is false → Option (c).

Question 5

Two resistors of resistances R₁ = 4 Ω and R₂ = 12 Ω are connected in parallel. What is the equivalent resistance of the combination?

Accepted Answer

Option (C) is correct. Explanation: For resistors in parallel, the equivalent resistance R_eq is given by: 1/R_eq = 1/R₁ + 1/R₂ → 1/R_eq = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 ∴ R_eq = 12/4 = 3 Ω

Question 6

The resistivity of a metallic conductor increases with rise in temperature. This is because, with rise in temperature:

Accepted Answer

Option (B) is correct.

Explanation: Resistivity is given by ρ = m/(ne²τ), where τ is the relaxation time. With rise in temperature, lattice ions vibrate more vigorously, increasing the frequency of electron–ion collisions and thus decreasing the relaxation time τ. Since ρ ∝ 1/τ, resistivity increases. The number density n, mass m, and charge e of electrons remain essentially unchanged with temperature.

Question 7

The resistivity of a metallic conductor increases with rise in temperature. This is because, with increase in temperature, the

Accepted Answer

Option (B) is correct.

Explanation: Resistivity is given by ρ = m / (n e² τ), where τ is the relaxation time. As temperature rises, lattice ions vibrate more vigorously, causing more frequent collisions of free electrons with the lattice. This decreases the relaxation time τ, and since ρ ∝ 1/τ, the resistivity increases.

Question 8

A wire of resistance R is stretched uniformly so that its length becomes double. What is the new resistance of the wire?

Accepted Answer

Option (d) is correct.

Explanation: Using R = ρL/A, when a wire is stretched to double its length, its volume (V = AL) remains constant.

Since A·L = A'·L' → A·L = A'·(2L) → A' = A/2.

∴ New resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R.

Question 9

A metallic conductor carries a steady current. If the number density of free electrons in the conductor is n, their drift velocity is v_d, charge on electron is e, and cross-sectional area is A, which one of the following correctly expresses the electric current I through the conductor?

Accepted Answer

Option (B) is correct.

Explanation: By the relation between drift velocity and current, the current through a conductor is given by I = nAevd, where n is the number density of free electrons, A is the cross-sectional area, e is the magnitude of electron charge, and vd is the drift speed. This follows from the fact that in time Δt, all electrons in the volume AΔx = A(vdΔt) cross a given section, giving charge ΔQ = nAevdΔt, and hence I = ΔQ/Δt = nAevd.

Question 10

The resistance of a tungsten filament at 20°C is 20 Ω. Find the resistance of the filament at 120°C. (Temperature coefficient of resistance of tungsten at 20°C is 4.5 × 10⁻³ K⁻¹.)

Accepted Answer

The resistance of a conductor at temperature T is given by:

RT = R0 [1 + α (T − T0)]

where R0 = resistance at reference temperature T0, and α = temperature coefficient of resistance.

Here: R0 = 20 Ω, T0 = 20°C, T = 120°C, α = 4.5 × 10−3 K−1

Substituting:

RT = 20 × [1 + (4.5 × 10−3) × (120 − 20)]

RT = 20 × [1 + (4.5 × 10−3) × 100]

RT = 20 × [1 + 0.45]

RT = 20 × 1.45

∴ RT = 29 Ω

Question 11

Two cells, each of emf ε and internal resistance r, are connected in parallel to an external resistance R. Using Kirchhoff's laws, find an expression for the current through R.

Accepted Answer

Let the current supplied by each cell be I₁ and I₂, and the current through R be I.

By Kirchhoff's Current Law (KCL): I = I₁ + I₂

Applying Kirchhoff's Voltage Law (KVL) to the loop containing cell 1 and R:
ε − I₁r − IR = 0  … (1)

Applying KVL to the loop containing cell 2 and R:
ε − I₂r − IR = 0  … (2)

From (1) and (2): I₁ = I₂ (since both cells are identical)
∴ I = 2I₁

From equation (1): ε − (I/2)r − IR = 0
→ ε = I(r/2 + R)

∴ I = ε / (R + r/2)

Question 12

A nichrome wire of length 2 m and cross-sectional area 0·5 mm² has a resistance of 4·4 Ω. Calculate the resistivity of nichrome. Also state how resistivity of a metallic conductor changes with rise in temperature.

Accepted Answer

The resistivity ρ of a material is related to resistance R, length L, and cross-sectional area A by:

R = ρL/A  →  ρ = RA/L

Given: R = 4·4 Ω, A = 0·5 mm² = 0·5 × 10⁻⁶ m², L = 2 m

Substituting:

ρ = (4·4 × 0·5 × 10⁻⁶) / 2

∴ ρ = 1·1 × 10⁻⁶ Ω m

Effect of temperature: For a metallic conductor, resistivity increases with rise in temperature, as ρ = ρ₀(1 + αΔT), where α > 0 for metals (increased lattice vibrations cause more frequent collisions of free electrons).

Question 13

A nichrome wire of length 2 m and cross-sectional area 0·5 mm² has a resistance of 4·4 Ω. Calculate the resistivity of nichrome.

Accepted Answer

The resistivity of a material is given by:

R = ρL/A  →  ρ = RA/L

Given: R = 4·4 Ω, L = 2 m, A = 0·5 mm² = 0·5 × 10⁻⁶ m²

Substituting:

ρ = (4·4 × 0·5 × 10⁻⁶) / 2

ρ = 2·2 × 10⁻⁶ / 2

∴ ρ = 1·1 × 10⁻⁶ Ω m

Question 14

In the circuit shown below, a battery of EMF 12 V and internal resistance 1 Ω is connected to three resistors. Resistors R₁ = 3 Ω and R₂ = 6 Ω are connected in parallel, and this parallel combination is connected in series with R₃ = 2 Ω and the battery. (a) Find the equivalent resistance of the external circuit. (b) Find the current through R₃. (c) Find the potential difference across the parallel combination (R₁ and R₂).

Accepted Answer

(a) Finding the equivalent external resistance:

R₁ and R₂ are in parallel:

1/R_parallel = 1/R₁ + 1/R₂ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6

∴ R_parallel = 2 Ω

This parallel combination is in series with R₃:

R_ext = R_parallel + R₃ = 2 + 2 = 4 Ω

∴ Equivalent external resistance = 4 Ω                    [1 mark]

(b) Finding the current through R₃:

By Ohm's law, the current delivered by the battery is:

I = EMF / (R_ext + r)

where r = 1 Ω is the internal resistance.

I = 12 / (4 + 1) = 12 / 5

∴ Current through R₃, I = 2.4 A                           [1 mark]

(c) Finding the potential difference across the parallel combination:

The potential difference across the parallel combination equals the terminal voltage minus the voltage drop across R₃:

V_parallel = I × R_parallel

V_parallel = 2.4 × 2

∴ Potential difference across parallel combination = 4.8 V  [1 mark]

Question 15

A technician is designing a battery-backup circuit for a small sensor module. Three resistors of resistances R₁ = 6 Ω, R₂ = 3 Ω and R₃ = 4 Ω are connected in a network along with a battery of emf 12 V and internal resistance r = 2 Ω. R₁ and R₂ are connected in parallel with each other, and this parallel combination is connected in series with R₃ and the battery.

(i) Find the equivalent resistance of the external circuit.
(ii) Find the current drawn from the battery.
(iii) Find the terminal voltage of the battery.
(iv) The technician now replaces R₃ with a wire of negligible resistance. Predict what happens to the terminal voltage of the battery and justify your answer.

Accepted Answer

(i) Equivalent resistance of external circuit:

For R₁ and R₂ in parallel:

1/R_parallel = 1/R₁ + 1/R₂ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6

∴ R_parallel = 2 Ω

This parallel combination is in series with R₃:

∴ R_ext = R_parallel + R₃ = 2 + 4 = 6 Ω

∴ Equivalent external resistance R_ext = 6 Ω                [1 mark]

(ii) Current drawn from the battery:

By Ohm's law, total resistance in circuit = R_ext + r = 6 + 2 = 8 Ω

Current I = E / (R_ext + r)

→ I = 12 / 8

∴ I = 1.5 A                [1 mark]

(iii) Terminal voltage of the battery:

The terminal voltage is given by:

V_terminal = E − I·r

→ V_terminal = 12 − (1.5)(2) = 12 − 3

∴ V_terminal = 9 V                [1 mark]

(iv) Effect on terminal voltage when R₃ is replaced by a wire of negligible resistance:

When R₃ = 0, external resistance = R_parallel = 2 Ω only.

New total resistance = R_parallel + r = 2 + 2 = 4 Ω

New current I′ = 12/4 = 3 A

New terminal voltage V′ = E − I′·r = 12 − (3)(2) = 12 − 6 = 6 V

The terminal voltage decreases (from 9 V to 6 V).

Reason: Because removing R₃ reduces the external resistance, which increases the current drawn from the battery. By the relation V_terminal = E − Ir, a larger current causes a greater potential drop (Ir) across the internal resistance, and so the terminal voltage falls.                [1 mark]

Question 16

Define resistivity (ρ) of a conductor. Write its SI unit. A wire of resistivity ρ is stretched uniformly so that its length becomes twice its original length. Show that the resistivity of the wire remains unchanged but its resistance becomes four times the original resistance.

Accepted Answer

Definition (½ mark):
Resistivity (ρ) of a conductor is defined as the resistance offered by a conductor of unit length and unit cross-sectional area. It depends only on the material and temperature of the conductor, not on its dimensions.
Formula: ρ = RA/L
SI unit: Ω m  (½ mark)

Why resistivity remains unchanged (½ mark):
Resistivity is an intrinsic property of the material. Stretching changes only the geometry (length and area) of the wire, not the nature of the material or its free-electron density. Therefore, ρ remains unchanged after stretching.

Derivation that resistance becomes four times (1½ marks):
Let the original wire have length L, cross-sectional area A, and resistance R.

Original resistance:
By the relation  R = ρL/A  …(i)

When the wire is stretched to twice its length, the volume of the wire remains constant.
Volume before stretching = Volume after stretching
→  A·L = A′·(2L)
→  A′ = A/2

New resistance after stretching:
R′ = ρ·(2L)/A′
→  R′ = ρ·(2L)/(A/2)
→  R′ = ρ·(2L)·(2/A)
→  R′ = 4·(ρL/A)
→  R′ = 4R    [using (i)]

∴ The resistance of the stretched wire is four times the original resistance, while the resistivity of the material remains unchanged.

Question 17

Three resistors of resistances 2 Ω, 3 Ω and 6 Ω are connected in parallel. This parallel combination is then connected in series with a battery of EMF 6 V and internal resistance 1 Ω. Calculate: (a) the effective resistance of the parallel combination, (b) the total current drawn from the battery, and (c) the terminal voltage of the battery.

Accepted Answer

(a) Effective resistance of the parallel combination:

For resistors in parallel, the equivalent resistance Rp is given by:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/2 + 1/3 + 1/6

1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1

∴ Rp = 1 Ω

(b) Total current drawn from the battery:

The total EMF of the battery is E = 6 V and internal resistance r = 1 Ω.
The parallel combination Rp = 1 Ω is in series with r.

By Ohm's law applied to the complete circuit:

I = E / (Rp + r)

I = 6 / (1 + 1)

I = 6 / 2

∴ I = 3 A

(c) Terminal voltage of the battery:

The terminal voltage V is the potential difference across the external (parallel) combination, given by:

V = E − I r

V = 6 − (3)(1)

V = 6 − 3

∴ V = 3 V

Question 18

A technician in a quality-control lab has a spool of nichrome wire of resistivity ρ = 1.10 × 10⁻⁶ Ω m and cross-sectional area A = 0.50 mm². She cuts a length L = 2.0 m from the spool and connects it across a cell of emf E = 3.0 V and internal resistance r = 0.5 Ω.

(i) Calculate the resistance R of the nichrome wire.
(ii) Find the drift velocity v_d of electrons in the wire when it carries current. (Given: number density of free electrons in nichrome n = 4.0 × 10²⁸ m⁻³, e = 1.6 × 10⁻¹⁹ C.)
(iii) The technician now doubles the length of the wire to 2L (same wire, same cross-section, same cell). Without doing a full calculation, predict and justify with a formula whether the drift velocity increases, decreases, or remains the same.
(iv) When the current flows through this wire, free electrons drift from the negative terminal side towards the positive terminal. Does this mean that every free electron in the wire moves in the same direction at all times? Justify your answer.

Accepted Answer

(i) State formula and calculate R:

By R = ρL/A,

A = 0.50 mm² = 0.50 × 10⁻⁶ m²

R = (1.10 × 10⁻⁶ × 2.0) / (0.50 × 10⁻⁶)

∴ R = 4.4 Ω

(ii) Find drift velocity v_d:

Current in circuit: I = E / (R + r)

I = 3.0 / (4.4 + 0.5) = 3.0 / 4.9

∴ I ≈ 0.612 A

By I = nAev_d,

v_d = I / (nAe)

v_d = 0.612 / (4.0 × 10²⁸ × 0.50 × 10⁻⁶ × 1.6 × 10⁻¹⁹)

v_d = 0.612 / (4.0 × 10²⁸ × 8.0 × 10⁻²⁶)

v_d = 0.612 / (3200)

∴ v_d ≈ 1.91 × 10⁻⁴ m s⁻¹

(iii) Effect of doubling the wire length:

When the length is doubled to 2L, the resistance becomes R′ = ρ(2L)/A = 2R = 8.8 Ω.

New current: I′ = E / (R′ + r) = 3.0 / (8.8 + 0.5) = 3.0 / 9.3 < I.

Since v_d = I / (nAe) and n, A, e are unchanged, a smaller current means a smaller drift velocity.

∴ The drift velocity decreases when the length is doubled (same cell, increased resistance reduces current and hence v_d).

(iv) Not all free electrons move in the same direction:

Free electrons in a conductor are in continuous random thermal motion with very high speeds (~10⁵ m s⁻¹). When a potential difference is applied, a very small net drift velocity (~ 10⁻⁴ m s⁻¹) is superimposed on this random motion in the direction opposite to the electric field E⃗.

At any instant, individual electrons move in all directions due to thermal agitation; only their average (net) displacement per unit time is directed from the negative to the positive terminal.

∴ No — not every free electron moves in the same direction at all times. The drift velocity represents the average velocity of the electron 'swarm', not the velocity of each individual electron.

Question 19

A technician is designing a circuit for a battery-testing station. A battery of EMF 12 V and internal resistance 2 Ω is connected to an external resistor R through a switch S. When S is closed, a current of 2 A flows through the circuit.

(i) Find the value of the external resistance R.

(ii) Calculate the terminal voltage of the battery when S is closed.

(iii) The technician replaces the battery with a second battery of the same EMF (12 V) but internal resistance 0.5 Ω. What current would flow if it is connected to the same external resistance R? Comment on which battery is more efficient for this application and why.

Accepted Answer

(i) Finding external resistance R:

By Ohm's law for a complete circuit, the EMF of a cell is related to current, external resistance, and internal resistance as:

E = I(R + r)

Substituting: 12 = 2 × (R + 2)

→ R + 2 = 6

→ R = 4 Ω

∴ External resistance R = 4 Ω

(ii) Terminal voltage of the battery:

The terminal voltage V is the potential difference across the external resistance (or equivalently, EMF minus the voltage drop across internal resistance):

V = E − Ir

Substituting: V = 12 − (2 × 2) = 12 − 4

∴ Terminal voltage V = 8 V

(iii) Current with the second battery (r′ = 0.5 Ω, R = 4 Ω):

Applying E = I′(R + r′):

12 = I′(4 + 0.5)

→ I′ = 12 / 4.5

∴ I′ = 2.67 A (approximately)

Comment on efficiency:

The efficiency of a cell is defined as the ratio of useful power delivered to the external resistance to the total power generated:

η = P_external / P_total = I²R / (I²(R + r)) = R / (R + r)

For Battery 1 (r = 2 Ω): η₁ = 4 / (4 + 2) = 4/6 ≈ 66.7%

For Battery 2 (r′ = 0.5 Ω): η₂ = 4 / (4 + 0.5) = 4/4.5 ≈ 88.9%

∴ Battery 2 (r = 0.5 Ω) is more efficient for this application because its lower internal resistance results in a smaller voltage drop across the cell itself, delivering a greater fraction of the total power to the external circuit.

Question 20

A technician working in an electronics lab has a cylindrical wire of resistivity ρ = 2.0 × 10⁻⁶ Ω m, length L = 2 m, and cross-sectional area A = 0.5 × 10⁻⁶ m². The wire is connected to a battery of emf E = 6 V and internal resistance r = 1 Ω.

(i) Calculate the resistance R of the wire and the current I drawn from the battery.

(ii) The technician now stretches this wire uniformly until its length becomes 3L (i.e., 3 times the original length). Assuming the volume of the wire remains constant during stretching, find the new resistance R′ of the stretched wire.

(iii) If the stretched wire (resistance R′) is now connected to the same battery (emf E = 6 V, internal resistance r = 1 Ω), find the new current I′ and the percentage decrease in current compared to the original current I.

(iv) The technician notices that the terminal voltage of the battery has changed after stretching. Explain why the terminal voltage increases when the external resistance increases, using the relation V = E − Ir.

Accepted Answer

(i) Finding R and I:

The resistance of a wire is given by:
R = ρL/A

Substituting values:
R = (2.0 × 10⁻⁶ × 2) / (0.5 × 10⁻⁶)
R = (4.0 × 10⁻⁶) / (0.5 × 10⁻⁶)
∴ R = 8 Ω

Current drawn from the battery:
By Ohm's law (including internal resistance): I = E / (R + r)
I = 6 / (8 + 1) = 6 / 9
∴ I = 0.67 A (= 2/3 A)

(ii) Finding new resistance R′ after stretching:

When the wire is stretched to length L′ = 3L, volume is conserved:
Volume = A × L = A′ × L′
A × L = A′ × 3L
∴ A′ = A/3

New resistance:
R′ = ρL′/A′ = ρ(3L)/(A/3) = 9 × (ρL/A) = 9R
R′ = 9 × 8
∴ R′ = 72 Ω

(iii) Finding I′ and percentage decrease:

I′ = E / (R′ + r) = 6 / (72 + 1) = 6 / 73
∴ I′ ≈ 0.082 A

Percentage decrease in current:
= [(I − I′) / I] × 100
= [(2/3 − 6/73) / (2/3)] × 100
= [(146/219 − 18/219) / (146/219)] × 100
= [128/219 ÷ 146/219] × 100
= (128/146) × 100
∴ Percentage decrease ≈ 87.7% ≈ 88%

(iv) Explanation of increased terminal voltage:

By Kirchhoff's voltage law, the terminal voltage of a battery is:
V = E − Ir

When the external resistance R increases, the current I = E/(R + r) decreases. Since the voltage drop across the internal resistance (= Ir) decreases, the terminal voltage V = E − Ir increases. Thus, more of the emf is available across the external circuit when a higher external resistance is connected.

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